number of ultrafilters

Theorem.

Let X be a set.
The number (http://planetmath.org/CardinalNumber) of ultrafilters (http://planetmath.org/Ultrafilter) on X is |X| if X is finite (http://planetmath.org/Finite),
and 22|X| if X is infinite (http://planetmath.org/Infinite).

Proof.

If X is finite then each ultrafilter on X is principal,
and so there are exactly |X| ultrafilters.
In the rest of the proof we will assume that X is infinite.

Let F be the set of all finite subsets of X,
and let Φ be the set of all finite subsets of F.

For each A⊆X define
BA={(f,ϕ)∈F×Φ∣A∩f∈ϕ},
and BA∁=(F×Φ)∖BA.
For each 𝒮⊆𝒫⁢(X)
define ℬ𝒮={BA∣A∈𝒮}∪{BA∁∣A∉𝒮}.

Let 𝒮⊆𝒫⁢(X),
and suppose A1,…,Am∈𝒮
and D1,…,Dn∈𝒫⁢(X)∖𝒮,
so that we have
BA1,…,BAm,BD1∁,…,BDn∁∈ℬ𝒮.
For i∈{1,…,m} and j∈{1,…,n}
let ai,j be such that
either ai,j∈Ai∖Dj or ai,j∈Dj∖Ai.
This is always possible, since Ai≠Dj.
Let f={ai,j∣1≤i≤m, 1≤j≤n},
and put ϕ={Ai∩f∣1≤i≤m}.
Then (f,ϕ)∈BAi, for i=1,…,m.
If for some j∈{1,…,n} we have Dj∩f∈ϕ ,
then Dj∩f=Ai∩f for some i∈{1,…,m},
which is impossible,
as ai,j is in one of these sets but not the other.
So Dj∩f∉ϕ,
and therefore (f,ϕ)∈BDj∁.
So (f,ϕ)∈BA1∩⋯∩BAm∩BD1∁∩⋯∩BDn∁.
This shows that any finite subset of ℬ𝒮
has nonempty intersection,
and therefore ℬ𝒮
can be extended to an ultrafilter 𝒰𝒮.

Suppose ℛ,𝒮⊆𝒫⁢(X) are distinct.
Then, relabelling if necessary, ℛ∖𝒮 is nonempty.
Let A∈ℛ∖𝒮.
Then BA∈𝒰ℛ,
but BA∉𝒰𝒮 since BA∁∈𝒰𝒮.
So 𝒰ℛ and 𝒰𝒮 are distinct
for distinct ℛ and 𝒮.
Therefore {𝒰𝒮∣𝒮⊆𝒫⁢(X)} is
a set of 22|X| ultrafilters on F×Φ.
But |F×Φ|=|X|,
so F×Φ has the same number of ultrafilters as X.
So there are at least 22|X| ultrafilters on X,
and there cannot be more than 22|X|
as each ultrafilter is an element of𝒫⁢(𝒫⁢(X)).
∎

Corollary.

Proof.

Let X be an infinite set.
By the theorem, there are 22|X| ultrafilters on X.
If 𝒰 is an ultrafilter on X,
then 𝒰∪{∅} is a topology on X.
So there are at least 22|X| topologies on X,
and there cannot be more than 22|X|
as each topology is an element of 𝒫⁢(𝒫⁢(X)).
∎